Answer:
At 81. 52 Deg C its resistance will be 0.31 Ω.
Explanation:
The resistance of wire =
Where 
=Resistance of wire at Temperature T
= Resistivity at temperature T ![=\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]](https://tex.z-dn.net/?f=%3D%5Crho_0%20%5C%20%5B1%20%5C%20%2B%20%5Calpha%5C%20%28T-T_0%5C%20%29%5D)
Where 
l=Length of the wire
& A = Area of cross section of wire
For long and thin wire the resistance & resistivity relation will be as follows
![\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}](https://tex.z-dn.net/?f=%5Cfrac%7B0.25%7D%7B0.31%7D%3D%5Cfrac%7B1%7D%7B%5B1%2B%5Calpha%28T-20%29%5D%7D)
T = 81.52 Deg C
The energy in electron volts of the photons that has the following frequencies is as follows:
- 620 THz = 2.564eV
- 3.10GHz = 1.28 × 10-⁵eV
- 46.0 MHz = 1.902 × 10-⁷eV
<h3>How to calculate energy?</h3>
The energy of a photon can be calculated using the following formula:
E = hf
Where;
- E = energy
- h = Planck's constant (6.626 × 10-³⁴ J/s)
- f = frequency
First, we convert the frequencies to hertz as follows;
- 620THz = 6.2 × 10¹⁴Hz
- 3.10GHz = 3.1 × 10⁹Hz
- 46.0MHz = 4.6 × 10⁷Hz
- E = 6.626 × 10-³⁴ × 6.2 × 10¹⁴ = 2.564eV
- E = 6.626 × 10-³⁴ × 3.1 × 10⁹ = 1.28 × 10-⁵eV
- E = 6.626 × 10-³⁴ × 4.6 × 10⁷ = 1.902 × 10-⁷eV
Learn more about energy of a photon at: brainly.com/question/2393994
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